Chapter 35 Physical Optics: Light as a wave

Size: px

Start display at page:

Download "Chapter 35 Physical Optics: Light as a wave"

Myron Summers
6 years ago
Views:

1 Chapter 35 Physical Optics: Light as a wave Interference happens when two or more waves combine Examples of no-pigmented color presentation (iridescence): Beautiful oil spills Colors on soap bubble Color of some animal s skin or feather Some gemstones (opal) 1

2 Peacock speading its feathers. Every branch carries a series of bright colored spots, the actual color depending on the exact position of the spot on the branch and on the angle of the incident light. 2

3 Detail of a peacock feather. The iridescent plumage of the peacock originates in the fine side branches of the feathers More at 3

$These ordered silica spheres produce the internal colors by causing the interference and diffraction of light passing through the microstructure of opal (Klein and Hurlbut, 1985, p.$

4 Opal: composed of silica spheres lacks color pigments At the micro scale precious opal is composed of hexagonal or cubic closely packed silica spheres some 150 to 300 nm in diameter. These ordered silica spheres produce the internal colors by causing the interference and diffraction of light passing through the microstructure of opal (Klein and Hurlbut, 1985, p. 444) 4

5 Silica spheres are transparent Slow sedimentation in a long column. Sedimentation on a large flat surface SiO 2 spheres Diameter: nm Water 5

6 METAL INFILTRATED OPAL AND METALLIC REPLICA Dissolve SiO2 with HF High pressure melt-infiltration Metal Infiltrated Opal Porous Silica Opal Inverse Opal or Metallic Replica 6

$diffraction and$ interference

7 METALLIC INVERSE OPAL: Color is due to diffraction and interference Antimony Replica at different angles of illumination 7

8 Interference of the waves When two or more waves arrive at a point according to the superposition principle the net result is a linear sum of the wavefunctions of the two waves. We use mathematical or graphical techniques to find the resulting wavefunction. Phasor s are graphical tools for representing the wavefunctions and their interactions. 8

9 Interference and coherent sources Interference is overlap of two or more waves in space and it is governed by principle of superposition. Principle of superposition: when two or more waves overlap, the resultant disturbance at any point and any instant is found by adding the instantaneous disturbances that would be produced at that point by the individual waves. For electromagnetic waves the disturbance is amplitude of the electric or magnetic field components. 9

10 We need coherent sources to observe interference patterns Why? 10

$Monochromatic sinusoidal wave Example: waves on surface of a pond Speed of the wave is the same in all directions No refraction Wavefronts (surfaces of$

11 Monochromatic sinusoidal wave Example: waves on surface of a pond Speed of the wave is the same in all directions No refraction Wavefronts (surfaces of constant phase): concentric circles Monochromatic: single wavelength of single frequency sinusoidal wave Example of monochromatic waves: Lasers (almost). 11

12 Constructive & destructive interference I Identical sources S 1 and S 2 same wavelengths and amplitudes: λ 1 = λ 2, A 1 =A 2, Coherent sources: λ 1 = λ 2 and in phase or synchronized with constant phase relationship (special case φ 1 = φ 2 ), same polarization if the waves are transverse (electromagnetic waves) Example for identical sources: two identical parallel rods (antenna) emitting EM waves 12

13 Constructive & destructive interference II Constructive Interference; travel time from S 1 and S 2 to a is equal; waves arrive in-phase Travel time from S 1 and S 2 to b or c is not equal; Constructive Interference: waves arrive in phase if the travel time difference correspond to integer number of wavelengths. Destructive Interference: waves arrive out of phase if the travel time difference correspond to half-integer number of wavelengths 13

14 Constructive & destructive interference III In general for sources in-phase or with constant phase relationship: Constructive interference happens if: r 2 r 1 = mλ (m = 0,±1,±2,...) integer number of λs Destructive interference happens if: r 2 r 1 = ( m + 1 / 2)λ (m = 0, ±1, ±2,...) half-integer number of λs If the amplitudes are equal then with half-λ phase difference, waves cancel each other. r 1 and r 2 are the distances of point of observation from the sources. 14

15 Anti-nodal curves: denoting constructive interference Nodal curves: denoting destructive interference nodal curve 15

16 Questions Standing waves and interference patterns are not the same What are the similarities? What are the differences? Where do we find two identical light sources? How can we produce a two source interference pattern? 16

17 Thomas Young s experiment (1800) 17

18 Thomas Young s experiment (1800) m 18

19 Thomas Young s experiment (1800) Light path difference between the rays arriving at y m : r 2 r 1 = d sinθ Condition for a bright fringe: r 2 r 1 = d sinθ = mλ; (m = 0,±1,±2,...) Condition for a dark fringe: r 2 r 1 = d sinθ = ( m + 1/ 2)λ; (m = 0,±1,±2,...) Position of a fringe on the screen: y m = R tanθ m tanθ m sinθ m for small angles or when d << R and then θ θ m y m = R tanθ m Rsinθ m Position of a bright fringe in Young's experiment: y m,bright = R mλ d ( Position of a bright fringe in Young's experiment: y m,dark = R m + 1/ 2 )λ d 19

20 Example In a Young s double slit experiment the slit width is 45 micron, the spacing between the slits is 0.2 mm, and the screen is located at m. Calculate position of the second order bright and dark fringe if wavelength of the light is 633 nm (red) (answer: y 2,bright,red =13 mm, y 2,dark,red =16 mm) If we change wavelength of the source to 532 nm (green), how the fringe spacing will change? (answer: y 2,bright,red =11 mm, y 2,dark,red =13 mm) Application: can we use a double slit to separate colors? 20

21 Applications of Interference in fields of science & engineering Chemistry / biology / Spectroscopy Biology / Biomedical engineering Mechanical Opto-mechanical Electronics / Optoelectronics / Electro-optics Material science engineering / Metrology Telecommunication / Military / Radar Aerospace / remote sensing / filter design Photography / Imaging / filtering 21

22 Targeted radio transmitter design An application of the interference Identical vertical dipole antennas oscillating at 1.5 MHz Pairs of antenna used to direct emitted energy in a particular direction Dark fringe area: bad reception Bright fringe area: good reception 22

23 Phasors: an alternative way of representing a wavefunction Each sinusoidal function is shown as a rotating vector or phasor whose projection on the x axis shows the instantaneous value of the function. Phasor s length represents the amplitude of the wave Phasor s angle with the positive x direction represents the phase of the wave (argument of the cosine function). Using phasors as a tool we can study superposition of the waves y y y E E E ωt E(t) = E cos(ωt) x ωt + φ E(t) = E cos(ωt + φ) x ωt + φ E(t) = E cos(ωt) + E cos(φ) x 23

24 Intensity in interference patterns using Phasors E 1 (t) = E cos(ωt) E 2 (t) = E cos(ωt + φ) E p = E 1 + E 2 a vector sum of E 1 and E 2 Magntude of E p (from triangonomety) E p 2 = E 2 + E 2 2E 2 cos(π φ) E p 2 = E 2 + E 2 + 2E 2 cosφ = 2E 2 (1+ cosφ) Using 1+cosφ=2cos 2 (φ / 2) I E p 2 = 4E 2 cos 2 (φ / 2) Amplitude of two wave interference is independent of time: E p = 2E cos φ 2 depends only on the initial phase difference! 24

25 Intensity of interference patterns using superposition of two waves E p is Independent of time and frequency only depends on initial phase differences and amplitudes 25

26 Amplitude and phase difference E total = E p cos ωt + φ 2 = 2E cos φ 2 cos ωt + φ 2 When the initial phase difference is even integer of π : φ = 0, ± 2π ± 4π,...,±2mπ then the waves are in-pahse and resulting amplitude is maximum E p = 2E When the initial phase difference is an odd integer of π: φ = π,±3π,...,(2m + 1)π then the waves are out of pahse and resulting amplitude is E p = 0 Conclusion : superposition of two waves with same frequency and amplitude but different pahses yeild a wave with same frequency and an amplitude that varies between zero and twice the amplitude of the original waves. 0 E p 2E 26

27 Intensity in two source interference 27

28 Intensity distribution 28

29 Interaction of two waves What happens when two or more waves exist in one point of space? Based on what principle the waves interact? What is the result of interaction of two or more waves? What is the relationship of the interference pattern with the phases of the waves involved? 29

30 Young s double slit experiment Position of a bright fringe in Young's experiment: y m,bright = Rmλ / d Position of a dark fringe in Young's experiment: y m,dark = R m where (m = 0,±1,±2,...) Amplitude of the superimposed wave at point P: E total = E p cos ωt + φ 2 where E = 2E cos φ p 2 λ d Intensity in two source interference pattern: I = I 0 cos 2 φ 2 Where I 0 = 2ε 0 ce 2, and φ is the pahse difference between the waves arriving at point P from two sources. How to calculate the φ in a problem? 30

31 Intensity in two slit experiment 31

32 Phase, path & optical path Phase y = Asin ωt kx + φ ( ) At t = 0 phase constant or initial phase is = k wavenumber x Dis tan ce kx = 2π λ x = 2π x = 2π number of λs in the distance x λ Each λ is equivalent to 2π in phase. Oprical path = distance index of the environment OP = xn Optical path difference = path difference index OPD = Δxn = ΔOP or OPD = xδn + φ Phase phase difference = optical path difference wavenumber Δφ = Δxnk 32

33 Phase difference and path difference (problem geometry) 33

34 Example Use the relationship between phase and path difference to find the intensity of a double silt experiment as a function of path difference When the intensity is maximum? When it is minimum? Note: path difference is something we can measure in the lab easily. It is not so easy to measure phase difference. How we can measure phase difference between two waves? Answer: Work the problem

35 Interference in thin films Light rays reflected from two interfaces with path/ phase difference interfere. The condition for constructive interference is different for different wavelengths. As a result we see different colors at different angles 35

36 Interference at a wedge Path difference between two rays is almost equal to 2t (close to normal incidence) At the corner there is no path difference so we should see a bright fringe but experiment shows a dark fringe. What is the problem? 36

37 Reflected wave amplitude E r n a n b E i 37

38 Amplitude of reflected EM wave from an interface 38

39 Interference at a wedge Path difference between two rays is almost equal to 2t Wave 1 has zero phase shift due to reflection from the glass-air interface. Wave 2 has phase shift due to reflection from air-glass interface. That is why the left corner appears dark. Light path difference is zero but phase difference due to reflection is or half a cycle

40 General condition for constructive and destructive interference 40

41 Example: thin film interference Two microscope slides 10.0 cm long are in contact at one end and separate by a mm thick paper at the other end. What is the spacing of the interference fringes seen by reflection. Is the fringe at the line of contact a bright or dark one? Assume the slides are illuminated with monochromatic light with free space wavelength of 500 nm. (x m =mx1.25 mm) If n glass =1.52 and the gap is filled with water n water =1.33 what happens to the fringe spacing? (x=mx0.940mm) Now upper plate is a plastic n plastic =1.40, lower plate is a special glass with and n glass =1.6, and the gap is filled with silicon grease n grease =1.5. Is the fringe at contact point is a dark one or a bright one? 41

Non reflective or antireflection coatings (close to normal incidence) Conditions for antireflective coating: we have to have destructive interference between the rays 1 and 2: n coating <n glass t =

42 Non reflective or antireflection coatings (close to normal incidence) Conditions for antireflective coating: we have to have destructive interference between the rays 1 and 2: n coating <n glass t = λ / 4 The phase difference between rays 1 and 2 due to reflection is zero. Why? The phase difference due to path difference is Why? Total phase difference is so no light is reflected coating

43 Newton s rings; monochromatic source 43

Newton s rings with colored source 44 http://www.fas.harvard.

44 Newton s rings with colored source 44

45 Michelson interferometer Historically has been developed to measure effect of the Ether on the speed of light By moving M 2 back and forth a distance of, λ / 2 the light path changes by amount. λ By moving M 2 back and forth a distance of y, m fringes pass by the crosshair of the eyepiece. y = m λ 2 m = 2y λ Nowadays it is used for measuring wavelength, index of refraction, glass inspection, stress analysis 45

46 Summary 46

47 47

Some properties of waves: Huygens principle Superposition Coherence Interference Young s double-slit experiment Thin-film interference

Some properties of waves: Huygens principle Superposition Coherence Interference Young s double-slit experiment Thin-film interference Phys 2435: Chap. 35, Pg 1 Geometrical Optics Assumption: the dimensions